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patriot [66]
3 years ago
6

Mrs. Nogaki uses constant force to push a Costco shoping cart down the long store

Physics
1 answer:
Liono4ka [1.6K]3 years ago
7 0
Figure B. The second option.
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A car speeds up from 12.0 m/s to 16.0 m/s in 8.0 s. the acceleration of the car is?
maria [59]

Vi=12m/s Vf=16m/s t=8s a=? a=Vf-Vi/t=16-12/8=4/8=1/2 a=0.5m/s^2

6 0
3 years ago
Buhrs atomic model differed from ruthofords because it explained that
Scrat [10]

Buhrs atomic model differed from ruthofords because it explained that electrons exist in specified energy levels surrounding the nucleus. This means that, Ruthoford believed that electrons can't do very much. However, Buhrs' model showed that electrons are much more powerful than anyone else believes they can be.

6 0
3 years ago
Read 2 more answers
A 878-kg (1940 lb) dragster, starting from rest, attains a speed of 25.9 m/s (57.9 mph) in 0.62 s. (a) Find the average accelera
salantis [7]

Answer:

41.8m/s^2

Explanation:

Since the dragster starts from rest, initial velocity (u) = 0m/s, final velocity (v) = 25.9m/s, time (t) = 0.62s

From the equations of motion, v = u + at

a = (v - u)/t = (25.9 - 0)/0.62 = 25.9/0.62 = 41.8m/s^2

7 0
3 years ago
A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t
ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement  

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement  

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0
3 years ago
If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t
Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to  Biot-Savart law, A magnetic field generated by a current  carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

 μ₀ =permeability of free space  4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 =  μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50  X 10 ^-3 X 2π  X  X 12.8 X 10^-2/  4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0
3 years ago
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